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Chapter 6: Problem 17

Find the angle between \(\mathbf{v}\) and \(\mathbf{w} .\) Round to the nearest tenth of a degree. $$\mathbf{v}=2 \mathbf{i}-\mathbf{j}, \quad \mathbf{w}=3 \mathbf{i}+4 \mathbf{j}$$

Short Answer

Expert verified
The angle between the given vectors is about \(78.5^\circ\) when rounded to the nearest tenth of a degree.

Step by step solution

01

Calculation of Dot Product between V and W

We need to find the dot product between the vectors \(\mathbf{v}\) and \(\mathbf{w}\). The dot product of two vectors can be found by multiplying their corresponding \(i\) and \(j\) components and then summing up those results. For vectors \(\mathbf{v}=2 \mathbf{i}-\mathbf{j}\) and \(\mathbf{w}=3 \mathbf{i}+4 \mathbf{j}\), their dot product can be found as follows: \(\mathbf{v}\cdot\mathbf{w}=(2)(3)+(-1)(4)=6-4=2.\)
02

Calculation of Magnitudes of V and W

We calculate the magnitudes of both vectors. The magnitude (or length) of a vector can be found using the Pythagorean Theorem. For our vectors \(\mathbf{v}=2 \mathbf{i}-\mathbf{j}\) and \(\mathbf{w}=3 \mathbf{i}+4 \mathbf{j}\), the magnitudes can be found as follows: \(||\mathbf{v}||=\sqrt{(2)^2+(-1)^2}=\sqrt{5}\), \(||\mathbf{w}||=\sqrt{(3)^2+(4)^2}=\sqrt{25}=5.\)
03

Substitute in the Angle Formula

We substitute the results obtained in Steps 1 and 2 into the formula for finding the angle between two vectors: \(\cos \theta = \frac{\mathbf{v}\cdot\mathbf{w}}{||\mathbf{v}||~||\mathbf{w}||} = \frac{2}{\sqrt{5}*5}.\)
04

Calculate the Angle

The above expression gives us the cosine of the angle between the two vectors. We can find the angle itself by taking the arccosine (inverse cosine) of this value, rounding to the nearest tenth of a degree: \(\theta = \arccos(\frac{2}{\sqrt{5}*5}) \approx 78.5^\circ.\)

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Most popular questions from this chapter

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